Question

If sin 3a = cos(a-10) where 3a and (a-10) both are acute angles, then find the value of a

Answer 1

sin3a=cos(a-10)

sin3a=sin(90-a+10)

3a=100-a

4a=100

a=25degree

hope it is the answer.

Answer 2

Hey mate

answer

$the \: formula \: between \: \\ \sin( \alpha ) \: and \: \cos( \beta ) \: is \\ \sin( \alpha ) = \cos(90 - \beta ) \\ \sin( \alpha ) = \cos( \beta ) \\ so \: in \: the \: question \\ \sin(3a) = \cos(a - 10) \\ = > \sin(3a) = \cos(90 - (a - 10) ) \\ = > \sin(3a) = \cos(90 - a + 10) \\ = > \sin(3a) = \sin(100 - a) \\ = > (3a) = 100 - a \\ = > 3a + a = 100 \\ = > 4a = 100 \\ = > a = \frac{100}{4} = 25 \: degree$

the value of "a " is 25°

✨✨hope it helps✨✨